✔ 最佳答案
(1+x)²ⁿ = (1+x)ⁿ * (x+1)ⁿ = ∑(r=0,n) C(n,r)x ͬ * ∑(k=0,n) C(n,k)xⁿ⁻ᵏ … (i)
Set x= 1 to get
2²ⁿ = ∑(r=0,n) C(n,r) * ∑(k=0,n) C(n,k)
Expand RHS for 2²ⁿ = ∑(r=0,n)C(n,r)² + 2∑(r=0,n), k=0,n) (r≠k) C(n,r)C(n.k) … (ii)
The last term (2S) is what you want
To find ∑(r=0,n)C(n,r)² equate coefficients of xⁿ in (i), selecting combinations where r=k
C(2n,n) = ∑(r=k) C(n,r) C(n,k) = ∑(r=0,n) C(n,r)²
From (ii) : 2²ⁿ = C(2n,n) + 2S ⟹ S = ½( 2²ⁿ−C(2n,n) )